2023 H2 Paper 1_Solutions_Students

2023 A Level Paper 1 Suggested Solutions 1 Find the exact equation of the tangent to the curve ( ) 2 ln 11 5yx=− at the point where 2.x= [5] Solutions ( ) ( )( ) ( ) 2 ln 11 5 Differentiating with respect to : 1d 2 11 5 5 10 11 5 Equa d dAt 2, e, 10e d t 2 e 10e t ( 2) 10e 21e ion of the tangent a yx x y xxyx yxy x x yx yx =− = − − =− − = = =− = − =− − =− + 2 The first four terms of a sequence are given by 1u = 10, 2u = 61, 3u = 206 and 4u = 469. It is given that nu is a cubic polynomial. (a) Find nu in terms of n. [3] Solutions Since nu is a cubic polynomial, 32 nu an bn cn d= + + + for some a, b, c, d  ℝ. ( )1 10 1u a b c d= + + + = −−− ( )2 8 4 2 61 2u a b c d= + + + = −−− ( )3 27 9 3 206 3u a b c d= + + + = −−− ( )4 64 16 4 469 4u a b c d= + + + = −−− Using GC to solve (1), (2), (3) and (4), a = 4, b = 23, c = – 46, d = 29 324 23 46 29nu n n n = + − +

2023 A Level Paper 1 Suggested Solutions (b) Find the range of values of n for which nu is greater than 25 000. [2] Solutions 324 23 46 29nu n n n= + − + > 25 000 From GC, 16.9n i.e. 17n 3 Vectors a and b are such that ab = –1. It is also given that ( ab + a) is perpendicular to (ab + b). (a) Show that |ab| = 1. [3] Solutions Given that ( )a b a+ is perpendicular to ( )a b b+  ( ) ( ).0a b a a b b +  + = ( ) ( ) ( ) ( ) 2 . . . . 0 0 0 . 0 a b a b a b b a a b a b a b a b   +  +  + =  + + + = [since ab is perpendicular to both a and b ] 2 2 0 0 1 0 1 1 since 0 (shown) ab ab a b a b  + + − = =  =   y y = 25000 y = 4x3 + 23x2 – 46x + 29 16.86920 x

2023 A Level Paper 1 Suggested Solutions (b) Hence find the angle between the direction of a and the direction of b. [3] Solutions Let the angle between the direction of a and the direction of b be  . ( ) ( ) ( ) ( ) sin 1 1 cos 1 2 1 : tan 12 a b a b a b a b     = = −−−  = =− −−− =− Since tan 0 is obtuse and lies in the second quadrant. 1180 tan 1 135 − = − − =  4 (a) Find cos cos dpx qx x , where p and q are constants such that p  q and p  – q. [2] Solutions ( ) ( ) ( ) ( ) cos cos d 1 cos cos d2 sin sin1 ,2 where is an arbitrary constant px qx x p q x p q x x p q x p q x Cp q p q C = + + −  + − = + + +−   (b) Given that n  0, show that 2 sin coscos d x nx nxx nx x c nn= + + , where c is an arbitrary constant. [3] Solutions 2 sin sincos d d sin cos (shown) where is an arbitrary constant nx nxx nx x x x nn x nx nx cnn c =− = + + 

2023 A Level Paper 1 Suggested Solutions (c) Using the result in part (b) show that, for all positive integers n, the value of π 0 cos dx nx x can be expressed as 2 k n , where the possible value(s) of k are to be determined. [2] Solutions   ( ) ( ) π π 20 0 22 22 22 2 sin coscos d πsin π cos π cos00 cos π10 sin π 0 for all p